Question

16．如图，设OP与x轴的正方向的夹角为α，OP'与OP的夹角为β，现将OP绕O点旋转到与OP'重合，旋转角β=$\frac{π}{6}$，则这个旋转变换对应的矩阵为$[\begin{array}{l}{\frac{\sqrt{3}}{2}}&{-\frac{1}{2}}\\{\frac{1}{2}}&{\frac{\sqrt{3}}{2}}\end{array}]$．

Answer 1

分析 根据矩阵的线性变换，即可求得旋转变换对应的矩阵．

解答 解：由二阶矩阵的线性变换性质可得：$[\begin{array}{l}{cosβ}&{-sinβ}\\{sinβ}&{cosβ}\end{array}]$$[\begin{array}{l}{x}\\{y}\end{array}]$=$[\begin{array}{l}{xcosβ-ysinβ}\\{xsinβ+ycosβ}\end{array}]$，
则旋转变换对应的矩阵A=$[\begin{array}{l}{cosβ}&{-sinβ}\\{sinβ}&{cosβ}\end{array}]$=$[\begin{array}{l}{\frac{\sqrt{3}}{2}}&{-\frac{1}{2}}\\{\frac{1}{2}}&{\frac{\sqrt{3}}{2}}\end{array}]$，
故答案为：$[\begin{array}{l}{\frac{\sqrt{3}}{2}}&{-\frac{1}{2}}\\{\frac{1}{2}}&{\frac{\sqrt{3}}{2}}\end{array}]$．

点评 本题考查矩阵的线性变换的简单应用，考查转化思想，属于基础题．